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2 Criterion for entire function requires $f$ to be real-analytic, not just $C^\infty$; deleted 17 characters in body
You should keep in mind analytic continuation. Let $f:\mathbb R\rightarrow\mathbb R$ be a $C^\infty$ function ; if there exists an entire function extending $f$, then
(2) The function $f$ must be real-analytic.
Of course (2) is not sufficient: think about the real-analytic $x\mapsto\frac{1}{1+x^2}$, which does NOT have an entire extension, since by analytic continuation, that extension should coincide with $\mathbb C\ni z\mapsto \frac{1}{1+z^2}$, which has poles at $\pm i$. For a $C^\infty$ real-valued function real-analytic $f$ on the real line, f$, you can formulate a criterion dealing with the size of derivatives. Such a function has an entire extension iff $$\forall R>0,\exists C_R,\forall n\in\mathbb N,\quad \vert f^{(n)}(0)\vert\le C_R\frac{n!}{R^n}.$$ On the other hand, real-analyticity of a$C^\inftyf$on the real line is equivalent to $$\forall x\in \mathbb R,\exists r>0,\exists C>0, \exists R>0,\forall n\in\mathbb N,\quad \Vert f^{(n)}\Vert_{L^\infty(B(x,r))}\le C\frac{n!}{R^n}.$$ 1 You should keep in mind analytic continuation. Let$f:\mathbb R\rightarrow\mathbb R$be a$C^\infty$function ; if there exists an entire function extending$f$, then (1) It is unique by analytic continuation, (2) The function$f$must be real-analytic. Of course (2) is not sufficient: think about the real-analytic$x\mapsto\frac{1}{1+x^2}$, which does NOT have an entire extension, since by analytic continuation, that extension should coincide with$\mathbb C\ni z\mapsto \frac{1}{1+z^2}$, which has poles at$\pm i$. For a$C^\infty$real-valued function$f$on the real line, you can formulate a criterion dealing with the size of derivatives. Such a function has an entire extension iff $$\forall R>0,\exists C_R,\forall n\in\mathbb N,\quad \vert f^{(n)}(0)\vert\le C_R\frac{n!}{R^n}.$$ On the other hand, real-analyticity of a$C^\inftyf\$ on the real line is equivalent to $$\forall x\in \mathbb R,\exists r>0,\exists C>0, \exists R>0,\forall n\in\mathbb N,\quad \Vert f^{(n)}\Vert_{L^\infty(B(x,r))}\le C\frac{n!}{R^n}.$$